Question

Frank kicks a soccer ball off the ground and in the air with an initial velocity of 30 feet per second. Using the formula H(t) = -16t^2 + vt + s, what is the maximum height the soccer ball reaches?
14.1 feet
13.7 feet
13.2 feet
15.2 feet

Answer 1

1 don't know what 1'm doing wrong 1 keep getting 0.

Answer 2

@asnaseer

Answer 3

What is the "v" and "s" in the formula for H(t)?

Answer 4

velocity and speed 1 think.

Answer 5

Is this all the information given to you in the question?

Answer 6

yes. You're supposed to use the quadratic formula

Answer 7

If "v" is supposed to be the initial velocity then you can re-write this as:\[H(t)=-16t^2+30t+s\]That still leaves the unknown "s" in this equation. Usually "s" is used to denote distance, but here you are give H(t) to represent the "distance" in terms of height. Is this question from a text book?

Answer 8

no the question is on my homework.

Answer 9

s would be 0

Answer 10

I am getting a lot of "lag" on this site right now - so sorry if I seem to be taking a long time to answer.

Why do you say s=0?

Answer 11

that's just how my teacher told me to set it up. s represents how far from the ground the starting point is.

Answer 12

Ok, so we actually have:\[H(t)=-16t^2+30t\]correct?

Answer 13

yes but the 0 is necessary for the quadratic formula

Answer 14

Why do you think you need to use the quadratic formula?

The quadratic formula is usually used to solve equations like:\[15t^2+12t-3=0\]Where as here you are being asked to find the maximum value of H given:\[H(t)=-16t^2+30t\]Have you studied calculus yet?

Answer 15

no

Answer 16

this is algebra 1 honors. 1'm in 8th grade.

Answer 17

Have you been taught about how to rewrite a quadratic in vertex form?

Answer 18

1'm not very good at it.

Answer 19

a(x - h)^2 + k right?

Answer 20

Ok - but it seems that you have been taught that method. I therefore think that you should use this method to solve this.

Yes - that is the vertex form

Answer 21

\[h=\frac{ -b }{ 2a }\] \[k= c- \frac{ 2b }{ 4a }\] ?

Answer 22

If you look at the equation given to you, we have:\[H(t)=-16t^2+30t\]So first step would be to factor out the 16 to give:\[H(t)=16\left(-t^2+\frac{30t}{16}\right)\]

Answer 23

we can now factor the terms inside the braces to get something like \((x-h)^2\)

can you see how to do that?

Answer 24

yes

Answer 25

It's called completing the square

Answer 26

okay

Answer 27

what do you get?

Answer 28

1 have no idea. sorry. 1'm horrible at math. : /

Answer 29

np - lets take a look at an example to help you...

Answer 30

lets say we have:\[x^2+\color{red}{8}x\]the first thing we would do is look at the coefficient of \(x\) and halve it. In this case we would halve \(\color{red}{8}\) to get \(\color{red}{4}\). so we then calculate what this equals:\[(x+\color{red}{4})^2\]

Answer 31

oh okay, 1 see

Answer 32

sorry - my internet went down

Answer 33

1ts fine

Answer 34

ok - if we now expand this we get:\[(x+\color{red}{4})^2=x^2+\color{red}{8}x+\color{blue}{16}\]so you can see if get an extra term of \(\color{blue}{16}\)

we therefore need to subtract this in order to get our original equation, i.e.:\[x^2+\color{red}{8}x=(x+\color{red}{4})^2-\color{blue}{16}\]

Answer 35

If we now go back to your equation, we have:\[H(t)=16\left(-t^2+\frac{30t}{16}\right)\]and we need to complete the square on:\[-t^2+\frac{30t}{16}\]

Answer 36

here we have to be careful because we have a negative sign on the \(t^2\) term

Answer 37

ok

Answer 38

so first we can rewrite this as:\[-\left(t^2-\frac{30t}{16}\right)\]so we need to complete the square on:\[t^2-\frac{30t}{16}\]understand so far?

Answer 39

yes

Answer 40

so, we you recall my explanation above, we need to halve the coefficient of \(t\) in this case and write:\[(t-\frac{15}{16})^2\]

Answer 41

because \(\frac{15}{16}\) is half of \(\frac{30}{16}\)

Answer 42

okay

Answer 43

if we now expand this we get:\[(t-\frac{15}{16})^2=t^2-\frac{30t}{16}+\color{blue}{\left(\frac{15}{16}\right)^2}\]to get our desired terms we need to subtract the blue term.

therefore:\[t^2-\frac{30t}{16}=(t-\frac{15}{16})^2-\color{blue}{\left(\frac{15}{16}\right)^2}\]

Answer 44

making sense?

Answer 45

yes

Answer 46

good. now we can go back to our original equation which was:\[H(t)=16\left(-t^2+\frac{30t}{16}\right)=16\left(-\left(t^2-\frac{30t}{16}\right)\right)\]\[=16\left(-\left((t-\frac{15}{16})^2-\color{blue}{\left(\frac{15}{16}\right)^2}\right)\right)\]\[=16\left(-(t-\frac{15}{16})^2+\color{blue}{\left(\frac{15}{16}\right)^2}\right)\]

Answer 47

If you look at this equation can you see what the maximum value would be?

Answer 48

we can rewrite this slightly better as:\[H(t)=16\left(\color{blue}{\left(\frac{15}{16}\right)^2}-\color{red}{\left(t-\frac{15}{16}\right)^2}\right)\]

Answer 49

since the terms inside the outer braces are both squared, the maximum must be when the red term is zero.

Answer 50

since the blue term is a constant and the red term varies with \(t\)

Answer 51

okay

Answer 52

so we get the maximum height when \(t=\frac{15}{16}\) (which would make the red term equal to zero)

and the value of the maximum would be:\[H_{max}=16\left(\color{blue}{\left(\frac{15}{16}\right)^2}\right)=\cancel{16}\times\frac{15\times15}{\cancel{16}\times16}=\frac{225}{16}\approx14.1\text{ feet}\]

Answer 53

geez, it seems like a lot of work for an answer.

Answer 54

Thank you so much!

Answer 55

I took extra time to ensure you understood each step

Answer 56

once you get good at completing the square then you should be able to work through these types of problems very quickly

Answer 57

yw :)